Question

The line $x\cos \alpha +y\sin \alpha =p$ will be a tangent to the conic $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$ if

• A. $p^2=a^2{\sin }^2\alpha +b^2{\cos }^2\alpha$
• B. $p^2=a^2+b^2$
• C. $p^2=b^2{\sin }^2\alpha +a^2{\cos }^2\alpha$
• D. None of these

Answer 1

Answer: (C)

$p^2=b^2{\sin }^2\alpha +a^2{\cos }^2\alpha$

$x\cos \alpha +y\sin \alpha =p$

$\Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{x^2}{a^2}+\frac{1}{b^2}{\left(\frac{p-x\cos \alpha }{\sin \alpha }\right)}^2=1$

$\frac{x^2}{a^2}+\frac{1}{b^2}\frac{p^2+x^2{\cos }^2\alpha -\left(2p\cos \alpha \right)x}{{\sin }^2\alpha }=1$

$\left(b^2{\sin }^2\alpha \right)x^2+p^2{a}^2+\left(a^2{\cos }^2\alpha \right)x^2-(2p\cos \alpha .a^2)x=a^2{b}^2{\sin }^{\alpha }$

$\Rightarrow (a^2{\cos }^2\alpha +b^2{\sin }^2\alpha )x^2-\left(2p{a}^2\cos \alpha \right)x+p^2{a}^2-a^2{b}^2{\sin }^2\alpha =0$

For tangent -

$\Rightarrow \left(2p{a}^2\cos \alpha \right)=4(a^2{\cos }^2\alpha +b^2{\sin }^2\alpha ).(p^2{a}^2-a^2{b}^2{\sin }^2\alpha )$

$\Rightarrow 4p^2{a}^4{\cos }^2\alpha =4p^2{a}^4{\cos }^2\alpha -4a^4{b}^2{\sin }^2\alpha {\cos }^2\alpha +4p^2{a}^2{b}^2{\sin }^2\alpha -4a^2{b}^4{\sin }^4\alpha$

$\Rightarrow 4a^4{b}^2{\sin }^2\alpha {\cos }^2\alpha =4p^2{a}^2{b}^2{\sin }^2\alpha -4a^2{b}^4{\sin }^4\alpha$

$\Rightarrow p^2=a^2{\cos }^2\alpha +b^2{\sin }^2\alpha$.